Let $$\vec a,\,\vec b$$  and $$\vec c$$ be non-zero vectors such that $$\left( {\vec a \times \vec b} \right) \times \vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\,\vec a.$$      If $$\theta $$ is the acute angle between the vectors $${\vec b}$$ and $${\vec c},$$  then $$\sin \,\theta $$  equals :

Solution :
Given $$\left( {\vec a \times \vec b} \right) \times \vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\,\vec a$$
Clearly $${\vec a}$$ and $${\vec b}$$ are non-collinear
$$\eqalign{ & \Rightarrow \left( {\vec a.\vec c} \right)\vec b - \left( {\vec b.\vec c} \right)\vec a = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\,\vec a \cr & \therefore \vec a.\vec c = 0{\text{ and }} - \vec b.\vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right| \Rightarrow \cos \,\theta = \frac{{ - 1}}{3} \cr & \therefore \sin \,\theta = \sqrt {1 - \frac{1}{9}} = \frac{{2\sqrt 2 }}{3}\,\,\,\left[ {\theta {\text{ is acute angle between }}\vec b{\text{ and }}\vec c} \right] \cr} $$